/* Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,
              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1
return
[
   [5,4,11,2],
   [5,8,4,5]
]*/

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void pathSumHelper(TreeNode *node, int &cur, vector<int> &v_cur, int sum, vector<vector<int> > &res) {
        cur += node->val;
        v_cur.push_back(node->val);
        if (cur == sum && !node->left && !node->right) {
            // found one leaf, store result
            res.push_back(v_cur);
        }
        // proceed to child elements
        if (node->left) pathSumHelper(node->left, cur, v_cur, sum, res);
        if (node->right) pathSumHelper(node->right, cur, v_cur, sum, res);
        // remove this element
        cur -= node->val;
        v_cur.pop_back();
    }
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        vector<vector<int> > res;
        if (!root) return res;
        vector<int> v_cur;
        int cur = 0;
        pathSumHelper(root, cur, v_cur, sum, res);
        return res;
    }
};
